calculate the moles of oxygen needed to react with 1.20 g of cigarette lighter gas, butane, C4H10 in combustion reaction

Answer 1

First, write a correctly balanced equation:

2C₄H₁₀ + 13O₂ ==> 8CO₂ + 10H₂O .. balanced equation

Next, find moles of butane present. Use molar mass of butane = 58.12 g / mol

1.20 g 2C₄H₁₀ x 1 mol / 58.12 g = 0.0206 mols

Next, use the stoichiometry (mole ratio) of the balanced equation to find moles O₂ needed:

00206 mols 2C₄H₁₀ x 13 mols O₂ / 2 mols 2C₄H₁₀ = 0.134 mols O₂ needed (to 3 sig.figs.)